In Shulman's 2008 paper 'Set Theory for Category Theory', he includes amongst the axioms of $\sf NBG$ the axiom of limitation of size. Being well known to imply the axiom of global choice, it seems to me that since $\sf ZFC+I$ doesn't imply global choice, this should be inconsistent with his later remark that $\sf ZFC+I$ has a model of $\sf NBG$ given by $V_{\kappa}$ for the sets, and $\operatorname{Def}(V_{\kappa})$ for the proper classes. I'm sure there is something here I don't quite understand, I would greatly appreciate some clarification.
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2$\begingroup$ That later remark does not say that ZFC+I implies global choice; it says that it provides a model for global choice. A well-order of $V_\kappa$ in the universe will feel like a global well-order if you think $V_\kappa$ is the whole universe. $\endgroup$– KP HartCommented Dec 8 at 13:41
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Care is needed here.
It is true that $\sf ZFC+I$ implies the consistency of $\sf NBG$. And it is true that $\sf ZFC+I$ does not prove Global Choice.
However, when we talk about consistency statements, namely "Theory $T$ proves that theory $S$ has a model", it is a weaker condition. Similarly, $\sf ZF+DC+LM$ (here $\sf LM$ stands for "every set of reals is Lebesgue measurable") proves the consistency of $\sf ZFC$, even if it refutes the Axiom of Choice.
So, what we are really saying here is that there is a model of $\sf NBG$ in which there is a Global Choice function. And since the model is a set in our universe of $\sf ZFC+I$, a choice function on the model can, potentially, be a good candidate for being a Global Choice function.
Nevertheless, the statement you seem to refer to is itself wrong. It is certainly consistent that inside $\operatorname{Def}(V_\kappa)$ there are no well-orders of $V_\kappa$. But it is true that we can find other ways around this. We can take $V_{\kappa+1}$ or we can restrict to $L_\kappa$ and $L_{\kappa+1}=\operatorname{Def}(L_\kappa)$, for our model of $\sf NBG$.
answered Dec 8 at 13:52
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$\begingroup$ Thanks for your answer! I'm still unclear on whether or not you're saying that inside $V_{\kappa+1}$, there is a well-ordering of $V_{\kappa}$? If so, how would one prove this? Also I'm unfamiliar with the $L_{\kappa}$ notation? $\endgroup$ Commented Dec 8 at 13:59
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$\begingroup$ Yes. That is what I'm saying. $V_{\kappa+1}$ is the set of all the subsets of $V_\kappa$, and since $V_\kappa\times V_\kappa\subseteq V_\kappa$, it also contains all the functions and relations on $V_\kappa$. Apply the $\sf C$ in your $\sf ZFC$ and you're done. As for the $L$ notation, this is Gödel's constructible universe which is constructed by taking the definable power set instead of the full power set at each stage. If you're unfamiliar with that, feel free to ignore it, consider it a blackbox, or learn more set theory to be more familiar with it! :-) $\endgroup$– Asaf Karagila ♦Commented Dec 8 at 14:02
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1$\begingroup$ Depending on your preexisting knowledge, Jech might be good. Or Enderton. Or Just & Weese. You can take a look on my website, there are lecture notes in set theory which briefly touch on $L$. After reading those, Jech would probably be a suitable option. $\endgroup$– Asaf Karagila ♦Commented Dec 8 at 18:35
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1$\begingroup$ Thanks for pointing out that mistake. If and when I revise the note I'll fix it. Is there a natural subset of $V_{\kappa+1}$ that we could take as the classes to obtain a model of NBG that isn't a model of Morse-Kelley? E.g. could we use $\mathrm{Def}(V_\kappa \cup \{x\})$ where $x$ is some fixed well-ordering of $V_\kappa$? $\endgroup$ Commented 2 days ago
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1$\begingroup$ @Mike: That should work. I'd say you probably also want to make sure that your well-ordering is not "too long". Since there will only be $\kappa$ many ordinals definable from $V_\kappa$, but $\kappa^+$ ways to well-order it, if you picked a well-ordering which is "too long" you might run into some non-obvious problems (although, again, I don't know enough off-hand to say that's definitely the case). But picking a well-ordering that is amenable is probably enough. Namely, you want $x\cap V_\alpha$ to be a well-ordering of $V_\alpha$ for all $\alpha<\kappa$. $\endgroup$– Asaf Karagila ♦Commented 2 days ago